# $\lim_{k\to\infty}\frac{1}{k}(\sum_{i=1}^k \sin(\omega i)\sin(\omega(i-j))=\frac{1}{2\pi} \int_{0}^{2 \pi}\sin (\omega x)\sin(\omega(x-j))dx, j\ge0$ [closed]

Is this a standard result with a well known proof? Or would anyone know how to go about proving it?

$$\lim_{k\to\infty}\frac{1}{k}(\sum_{i=1}^k \sin(\omega i)\sin(\omega(i-j))=\frac{1}{2\pi} \int_{0}^{2 \pi}\sin (\omega x)\sin(\omega(x-j))dx,$$ where $$j$$ is an integer $$\ge 0$$ and $$\omega \in \mathbb{R}$$.

I have shown that this seemingly works numerically for large values of $$k$$, but wanted to see if this is true analytically.

Using the Riemann sum: $$\lim_{n\to \infty}\sum_{i=1}^nf(x_i)\Delta x=\int_a^bf(x)dx$$ where $$\Delta x=\frac{b-a}{n}$$ and $$x_i=a+\Delta x \cdot i$$,
you get, $$\frac{1}{2\pi}\int_0^{2\pi}\sin(\omega x)\sin(\omega(x-j))dx=\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^{k}\sin\left(\omega\frac{2\pi i}{n}\right)\sin\left(\omega\left(\frac{2\pi i}{n}-j\right)\right)$$ and using trigonometric identities you should be able to show $$\frac{1}{k}\sum_{i=1}^{k}\sin\left(\omega\frac{2\pi i}{n}\right)\sin\left(\omega\left(\frac{2\pi i}{n}-j\right)\right)=\frac{1}{k}\sum_{i=1}^{k}\sin(\omega i )\sin\left(\omega i-j)\right)$$ yet, I was unable to achieve this.
• I don't believe this can be directly applied since in the original problem, $i$ are integers from $1$ through $k$, which are not on the interval $[0,2 \pi]$ Apr 2 at 17:17